Problem: ${\sqrt[3]{128} = \text{?}}$
Solution: $\sqrt[3]{128}$ is the number that, when multiplied by itself three times, equals $128$ First break down $128$ into its prime factorization and look for factors that appear three times. So the prime factorization of $128$ is $2\times 2\times 2\times 2\times 2\times 2\times 2$ Notice that we can rearrange the factors like so: $128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = (2\times 2\times 2) \times (2\times 2\times 2) \times 2$ So $\sqrt[3]{128} = \sqrt[3]{2\times 2\times 2} \times \sqrt[3]{2\times 2\times 2} \times \sqrt[3]{2}$ $\sqrt[3]{128} = 2\times 2 \times \sqrt[3]{2}$ $\sqrt[3]{128} = 4 \sqrt[3]{2}$